Question

When an ammeter of negligible internal resistance is inserted in series with circuit it reads $1A$. When the voltmeter of very large resistance is connected across $X$ it reads $1V$. When the point $A$ and $B$ are shorted by a conducting wire, the voltmeter measures $10V$ across the battery. The internal resistance of the battery is equal to

• A. Zero
• B. $0.5\Omega$
• C. $0.2\Omega$
• D. $0.1\Omega$

Answer 1

The correct option is C $0.2\Omega$

Initially ammeter measures current 1A through the circuit, when voltmeter measures voltage 1V the resistance is,
$R=\frac{V}{I}=1\Omega$
When the point A and B are shorted by a conducting wire, the voltmeter measures 10V across the battery.
Hence the current is
$I=\frac{10}{1}=10A$
Now, the voltage across the terminals of the battery is
$V=E-Ir$
$10=12-10r$
$10r=2$
$r=0.2\Omega$